python API url 级联生成

参考了一下公司 python 达人 rpc 接口级联 api 调用

rpc.api.users.list()

rpc.api.login(username='',password='')

rpc['api/users'](id=222)

写了一个 demo,python 确实很简洁,不到 30 行代码搞定

# coding=utf-8

class NameChain(object):
    def __init__(self,prefix,callback):
        self._prefix = prefix
        self._callback = callback

    def __getattr__(self,item):
        next_prefix = self._prefix + '/' + item
        return NameChain(next_prefix,self._callback)

    def __call__(self,**kw):
        self._callback(self._prefix,kw)

    def __getitem__(self,item):
        self._prefix = item
        return self

class Service(object):
    def __init__(self):
        self._api = NameChain('api',self._invoke)

    def _invoke(self,method,params):
        print '******************'
        print 'invoke method:',method

        if len(params) ==0:
            print 'params is empty'
            return

        print 'params is'
        for k,v in params.iteritems():
            print k,':',v

    @property
     def api(self):
         return self._api

s = Service()
api = s.api

# 调用方式一
print '调用方式一'
# api/users
api.users()   
# api/user    params {"id",111,"name":"leslie"}
api.user(id=111,name='leslie')
# pai/user/tags 
api.user.tags() 
 #api/user/tags/list params {"id",111,"session_id","love you forever"}
api.user.tags.list(id=111,session_id='love you forever')

# 调用方式二
print '调用方式二'
api['api/users']()
api['api/users'](id=111,name='leslie')
api['api/user/tags']()
api['api/user/tags/list'](id=111,session_id='love you forever')

 

posted @ 2015-05-21 22:58  LeslieFang  阅读(598)  评论(0)    收藏  举报